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zhenek [66]
3 years ago
14

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and

the fin is exposed to convective conditions characterized by h = 220 W/m2-K. It is reported that the fin efficiency is ηf = 0.65. Determine the fin length L and the fin effectiveness εf. Account for tip convection.
Physics
2 answers:
frozen [14]3 years ago
5 0

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

0.65=\dfrac{tanh37.08L}{37.08L}

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}

\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

svp [43]3 years ago
3 0

Answer:

The length of the fin is 35.2 mm

The effectiveness is 23.5

Solution:

As per the question:

Thermal conductivity, k = 160\ W/m-K

Diameter of the fin, D = 4 mm = 4\times 10^{- 3}\ m

Coefficient of convective heat transfer, h = 220\ W/m^{2}-K

Efficiency of the fin, \eta_{f} = 0.65

Now,

To calculate the length if the fin:

\eta_{f} = \frac{tanh\ mL_{c}}{mL_{c}}             (1)

where

L_{c} = L + \frac{D}{4} = (L + 1)\ mm

Also,

m = \sqrt{\frac{Ph}{A_{c}k}}

where

P = Perimeter of the fin = \pi D

A_{c} = Convection Area

A_{c} = \frac{\pi D^{2}}{4}

Now,

m = \sqrt{\frac{\pi Dh}{\frac{\pi D^{2}}{4}k}}

m = \sqrt{\frac{4h}{Dk}} = \sqrt{\frac{4\times 220}{4\times 10^{- 3}\times 160}} = 37.08\ m^{- 3}

Now, using eqn (1):

0.65 = \frac{tanh\ (37.08L_{c})}{37.08L_{c}}

tanh\ (37.08L_{c}) = 24.10L_{c}

Now, by trial-error:

L_{c} = 0.0362 m = 36.2\ mm

Also, we know that:

L_{c} = L + 1

36.2 = L + 1

L = 35.2 mm

Now, the fin effectiveness is given by:

\epsilon_{f} = \sqrt{\frac{Pk}{hA_{c}}}

\epsilon_{f} = \sqrt{\frac{\pi D\times 160}{220\times \frac{\pi D^{2}}{4}}}tanh\ mL_{c}

\epsilon_{f} = \sqrt{\frac{4\times 160}{220\times 4\times 10^{- 3}}}tanh\ (37.08\times 0.0362) = 23.5

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